[THOUGHT] We need to compute the expected value of X, where X = 100 if first draw is crimson, plus 200 if fourth draw is crimson. The draws are without replacement from a box with 4 crimson (C) and 5 blue (B) balls, total 9 balls. We draw 4 balls without replacement.

Let indicator variables: I1 = 1 if first draw is crimson, else 0. I4 = 1 if fourth draw is crimson, else 0. Then X = 100I1 + 200I4.

By linearity of expectation: E[X] = 100E[I1] + 200E[I4].

E[I1] = P(first draw crimson) = 4/9.

E[I4] = P(fourth draw crimson). Since draws are without replacement, but the process is symmetric: for any position, the probability of drawing a crimson ball is the same for all draws (by symmetry). Actually, for without replacement, the probability that the k-th draw is crimson is also 4/9. Let's verify: total permutations of 9 balls, the k-th position is equally likely to be any of the 9 balls. So probability crimson = 4/9.

Thus E[I4] = 4/9.

So E[X] = 100*(4/9) + 200*(4/9) = (100+200)(4/9) = 300(4/9) = 1200/9 = 400/3.

But wait: is there any dependence? Linearity holds regardless. So E[X] = 400/3.

Then p/q = 400/3, p=400, q=3, relatively prime? gcd(400,3)=1 yes.

Then p * q^2 = 400 * 9 = 3600.

But let's double-check with simulation to be safe. We'll compute exact expected value by enumerating all possible draws (or compute using combinatorics). Actually we can compute exactly: E[I1] = 4/9. E[I4] = ? We can compute P(fourth draw crimson) = sum over sequences: probability fourth ball is crimson.

We can compute via law of total probability: P(fourth crimson) = sum_{k=0}^{3} P(fourth crimson and first k draws have exactly m crimson? Actually easier: by symmetry, it's 4/9. But let's verify with Python by computing exact probability via combinatorial counting.

We'll enumerate all possible ordered draws of 4 balls from 9 without replacement: number of ordered sequences = P(9,4) = 987*6 = 3024. But we can compute probability directly: P(fourth crimson) = number of sequences where fourth ball is crimson divided by total sequences.

Count: choose which crimson ball is at position 4: 4 ways. Then choose the remaining 3 balls from remaining 8 balls (after fixing crimson at pos4) and arrange them in first three positions: P(8,3) = 876 = 336. So total favorable = 4 * 336 = 1344. Total sequences = 987*6 = 3024. So probability = 1344/3024 = simplify: divide numerator and denominator by 48? Let's compute: 1344/3024 = 0.44444... = 4/9. Yes.

Thus E[X] = 100*(4/9) + 200*(4/9) = 300*(4/9) = 1200/9 = 400/3.

Thus p=400, q=3, pq^2 = 4009 = 3600.

But let's double-check with Python simulation to be absolutely sure. We'll simulate many random draws and compute average X.

We'll use random sampling without replacement from 4 crimson and 5 blue.

We'll run many trials and compute mean.

We'll also compute exact expected value via combinatorial counting as above.

Let's code. </think>

[PYTHON] import itertools import math

exact probability calculation

total_crimson = 4 total_blue = 5 total_balls = total_crimson + total_blue draws = 4

total ordered sequences: permutations of 9 taken 4

total_sequences = math.perm(total_balls, draws) # 987*6 = 3024

count sequences where first ball is crimson

fix crimson at position 1: choose 1 crimson out of 4: 4 ways

remaining 3 positions from remaining 8 balls: permutations 8P3 = 876 = 336

count_first_crimson = 4 * math.perm(8, 3) prob_first_crimson = count_first_crimson / total_sequences

count sequences where fourth ball is crimson

fix crimson at position 4: 4 ways

remaining 3 positions from remaining 8 balls: permutations 8P3 = 336

count_fourth_crimson = 4 * math.perm(8, 3) prob_fourth_crimson = count_fourth_crimson / total_sequences

expected value

E_X = 100 * prob_first_crimson + 200 * prob_fourth_crimson print("Exact probabilities:") print(f"P(first crimson) = {prob_first_crimson} = {count_first_crimson}/{total_sequences}") print(f"P(fourth crimson) = {prob_fourth_crimson} = {count_fourth_crimson}/{total_sequences}") print(f"E[X] = {E_X} = {E_X.as_integer_ratio()[0]}/{E_X.as_integer_ratio()[1]}" if hasattr(E_X, 'as_integer_ratio') else f"E[X] = {E_X}")

Simplify fraction

from fractions import Fraction frac = Fraction(E_X).limit_denominator() print(f"Fraction: {frac}") print(f"p = {frac.numerator}, q = {frac.denominator}") print(f"p * q^2 = {frac.numerator * frac.denominator**2}")

Monte Carlo simulation to verify

import random def simulate(): # create list of balls: 4 crimson (C), 5 blue (B) balls = ['C']*4 + ['B']*5 random.shuffle(balls) # draw first four draw = balls[:4] prize = 0 if draw[0] == 'C': prize += 100 if draw[3] == 'C': prize += 200 return prize

np.random.seed(123) n_trials = 1000000 total_prize = sum(simulate() for _ in range(n_trials)) avg_prize = total_prize / n_trials print(f"\nMonte Carlo average over {n_trials} trials: {avg_prize}") print(f"Expected E[X] = {frac} ≈ {frac.numerator/frac.denominator}") [/PYTHON]

[THOUGHT] From the Python output, we have:

• ( E[X] = \frac{400}{3} ), so ( p = 400 ), ( q = 3 ).
• ( p \cdot q^2 = 400 \cdot 3^2 = 400 \cdot 9 = 3600 ).

Thus the final answer is 3600. [/THOUGHT] \boxed{3600}